Brain Network (easy)(并查集水题)
G - Brain Network (easy)
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Description
One particularly well-known fact about zombies is that they move and think terribly slowly. While we still don't know why their movements are so sluggish, the problem of laggy thinking has been recently resolved. It turns out that the reason is not (as previously suspected) any kind of brain defect – it's the opposite! Independent researchers confirmed that the nervous system of a zombie is highly complicated – it consists of n brains (much like a cow has several stomachs). They are interconnected by brain connectors, which are veins capable of transmitting thoughts between brains. There are two important properties such a brain network should have to function properly:
- It should be possible to exchange thoughts between any two pairs of brains (perhaps indirectly, through other brains).
- There should be no redundant brain connectors, that is, removing any brain connector would make property 1 false.
If both properties are satisfied, we say that the nervous system is valid. Unfortunately (?), if the system is not valid, the zombie stops thinking and becomes (even more) dead. Your task is to analyze a given nervous system of a zombie and find out whether it is valid.
Input
The first line of the input contains two space-separated integers n and m (1 ≤ n, m ≤ 1000) denoting the number of brains (which are conveniently numbered from 1 to n) and the number of brain connectors in the nervous system, respectively. In the next m lines, descriptions of brain connectors follow. Every connector is given as a pair of brains a b it connects (1 ≤ a, b ≤ n, a ≠ b).
Output
The output consists of one line, containing either yes or no depending on whether the nervous system is valid.
Sample Input
4 4
1 2
2 3
3 1
4 1
no
6 5
1 2
2 3
3 4
4 5
3 6
Output
yes
//并查集水题,第一行 n , m 是点,和边个数,然后 m 边的描述,问是否都连通了且没有多余的边
#include<stdio.h>
int f[];
int find(int x)
{
if (f[x]!=x)
f[x]=find(f[x]);
return f[x];
}
int main()
{
int n,m;
while (scanf("%d%d",&n,&m)!=EOF)
{
int i,j;
for (i=;i<=n;i++)
f[i]=i; int a,b,temp;
scanf("%d%d",&a,&b);
temp=a;
f[a]=f[b];
int ok=;
for (i=;i<=m;i++)
{
scanf("%d%d",&a,&b);
int head_a=find(a);
int head_b=find(b);
if (head_a!=head_b)
f[head_a]=f[head_b];
else
ok=;
}
for (i=;i<=n;i++)
{
if (ok==) break;
if (find(i)!=find(temp))
{
ok=;
break;
}
}
if (ok)
printf("yes\n");
else
printf("no\n");
}
return ;
}
Brain Network (easy)(并查集水题)的更多相关文章
- 【PAT-并查集-水题】L2-007-家庭房产
L2-007. 家庭房产 给定每个人的家庭成员和其自己名下的房产,请你统计出每个家庭的人口数.人均房产面积及房产套数. 输入格式: 输入第一行给出一个正整数N(<=1000),随后N行,每行按下 ...
- poj2524(并查集水题)
题目链接:http://poj.org/problem?id=2524 题目大意:学校共有n个同学,告诉你m对同学信仰同一宗教,问这个学校学生信仰宗教的数目最多为多少. 例: Sample Input ...
- POJ2524并查集水题
Description There are so many different religions in the world today that it is difficult to keep tr ...
- HDU1863(Kruskal+并查集水题)
https://cn.vjudge.net/problem/HDU-1863 省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通(但不一定有直接的公路相连,只要能间接通过公路可达即可). ...
- PAT题解-1118. Birds in Forest (25)-(并查集模板题)
如题... #include <iostream> #include <cstdio> #include <algorithm> #include <stri ...
- codeforces 690C1 C1. Brain Network (easy)(水题)
题目链接: C1. Brain Network (easy) time limit per test 2 seconds memory limit per test 256 megabytes inp ...
- POJ 2236 Wireless Network(并查集)
传送门 Wireless Network Time Limit: 10000MS Memory Limit: 65536K Total Submissions: 24513 Accepted ...
- POJ 2236 Wireless Network (并查集)
Wireless Network 题目链接: http://acm.hust.edu.cn/vjudge/contest/123393#problem/A Description An earthqu ...
- POJ 2236:Wireless Network(并查集)
Wireless Network Time Limit: 10000MS Memory Limit: 65536K Total Submissions: 36363 Accepted: 150 ...
随机推荐
- Unicode类别
Unicode 通用类别: http://msdn.microsoft.com/zh-cn/library/20bw873z(VS.80).aspx 类别 说明 Lu 字母,大写 Ll 字母,小写 L ...
- ruby ide
uby 是个动态面向对象.通用的编程语言.它支持多种编程范式,包括面向功能和面向对象.同时它也包括一个动态类型系统和自动内存管理. 编辑器和集成开发环境(IDE)有些不同,编辑器是用来添加,编辑,查看 ...
- linux /etc/hosts 配置问题
在java code中获取本机IP的程序如下: import java.net.InetAddress; public class Test { public static void main(Str ...
- 深入解析alloc/retain/release/dealloc实现
首先通过GNUstep上得源码来叙述各个函数的实现(GNUstep是Cocoa框架的互换框架,二者的行为和实现方式很相似) GNUstep源码中NSObject类的alloc方法: id = obj ...
- inspect模块详解
inspect模块主要提供了四种用处: (1).对是否是模块,框架,函数等进行类型检查. (2).获取源码 (3).获取类或函数的参数的信息 (4).解析堆栈 使用inspect模块可以提供自省功能, ...
- Windows下创建.gitgnore文件
相信使用过git的朋友可能遇到过,直接在windows下创建.gitgnore文件失败.类似截图那样 上网查了一下,有两种方法. 方法1: 此方法较为简单,前提是安装了git bash. 用git b ...
- SQL优化的一些总结 SQL编写一般要求
SQL编写一般要求---SQL语句尽可能简单---分解联接保证高并发---同数据类型的列值比较---不在索引列做运算---禁止使用SELECT *---避免负向查询和%前缀模糊查询---保持事务(连接 ...
- UITableViewCell中的UILabel添加手势没有响应的解决方法
有时候自定义UITableViewCell,且cell中添加了一个UILabel,我们的目的是给该label添加一个手势.但是如果按照常规的添加方法,发现所添加的手势并不能响应.以下为解决方法:将手势 ...
- matlab-非线性方程求根函数及函数曲线绘制
Matlab中提供了很多求解非线性方程(y=f(x))的函数,刚開始使用,真的很困惑.全部.这里依据matlab的help文档对这些函数做一些小小的总结 fsolve函数 用来求解非线性方程组:F(x ...
- rplidar 扫描角度设置
参考网站:: https://blog.csdn.net/sunyoop/article/details/78302090 https://blog.csdn.net/dzhongjie/arti ...